Saved Bookmarks
| 1. |
An ideal battery passes a current of 5A through a resistor . When it is connected to another resistance of 10 Omega in parallel, the current is 6A . Find the resistance of the first resistor. |
|
Answer» Solution :For IDEAL battery internal resistance r=0 Resistance of first wire= `R_1` Resistance of the second wire `R_2 = 10 OMEGA`  Current through `R_1 ` in the first case `i_1 = 5A` Current in the second case `i_2 = 6A` Effective resistance in the second case `R= (R_1 R_2)/(R_1 + R_2)` `V= I_1 R_1 and V= I_2 (R_1 R_2)/(R_1 + R_2)` `I_1 R_1 = I_2(R_1 R_2)/(R_1 + R_2)` `impliesI_1 = I_2 (R_2)/(R_1 + R_2)` `5 = 6 xx (10)/(R_1 + 10) implies5(R_1 + 10) = 60` `5R_1 + 50 = 60 , 5R_1 = 10` `R_1 = (10)/(5) = 2 Omega impliesR_1 = 2 Omega` |
|